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Shivansh Mathur
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How can we derive the formula for finding equivalent power in a series and parallel combination of 'n' resistors (with fixed resistance)?
Use that formula for the particular combination , in which a particular variable is common to all .Shivansh Mathur said:How can we derive the formula for finding equivalent power in a series and parallel combination of 'n' resistors (with fixed resistance)?
nR(series)=2xnR(parallel)Shivansh Mathur said:How can we derive the formula for finding equivalent power in a series and parallel combination of 'n' resistors (with fixed resistance)?
suranna said:nR(series)=2xnR(parallel)
Shivansh Mathur said:The same question is put up on this link -
https://answers.yahoo.com/question/index?qid=20080607083645AAqHHPw
Energy conservation principle. Where else would the input power come from or where would any surplus power go?Shivansh Mathur said:P(p) = P(1) + P(2) HOW?
Even this answer is too simple, I think. You cannot tell, without measurement, how an individual bulb will behave. The surface temperature will surely depend upon the actual filament and envelope design and not just on its power rating. I really don't like the idea of putting filament bulbs as 'examples' of Resistors - presumably in an attempt to make the topic more approachable so the original question is 'questionable'. Science teaching is full of dodgy examples, thought up by people who don't have a lot of practical experience or detailed knowledge of the example items they include in their question.Jony130 said:Well the answer is very simple;
If bulbs have 60W at 200V this means that the bulbs resistance is equal to:
P = V^2/R ---> R = V^2/P = 200V^2/60W = 667Ω and because we have three light bulbs connected in series the total resistance is equal to 3*667 = 2kΩ so the total power is equal to P = 200V^2/2kΩ = 20W This means that the book give you the correct equation.
But you should remember that this formula is only true for a device with a constant resistance. But this equation is not true for the light bulb. Becomes the light bulb resistance is not constant but will change with the apply voltage. See the graph of resistance of a 100W/230V light bulb in function of a supply voltages.
View attachment 87405
The resistance for 230V is equal 530Ω so current is equal 0.434A
But for 120V the current will not be equal to I = 120V/530Ω = 0.226A
As we can read form the graph the current will be equal I = 120V/ 380Ω = 0.315A
Thanks for asking the 1/P question. To be honest, I'd never thought of it in that way before.Shivansh Mathur said:Now I've got the answer and understood it completely.
no its notsophiecentaur said:Even this answer is too simple, I think. You cannot tell, without measurement, how an individual bulb will behave. .
It's halfway between a simple resistor and a real device. If you want to help someone get to the real world then your graph should be coupled with a caveat or they could take it as gospel. If you had added "one example of a 100W light bulb" I wouldn't have picked it up.William White said:no its not
from where did you get this pictureShivansh Mathur said:Sir, i myself have not done any math. I just seek to find a short way to calculate equivalent power in a circuit where devices of equal power rating are connected.
The source VOLTAGE is kept constant as you said.
for eg-
When three bulbs of 60W-200V rating are connected in series to a 200V supply, the power drawn by them will be : (Ans= 20W)
Now, i am asking for the proof of the formula that i have highlighted.
View attachment 87396
Have provided the circuit diagram also this time :)
View attachment 87397
Absolutely.berkeman said:The question still makes no sense.
There isn't a short way to get a reliable answer. Can you not accept that? The "short way" is only available for Resistors that do not change their value over their operating range. If you haven't done any Maths then I would advise you to avoid Electronic Engineering questions.Shivansh Mathur said:Sir, i myself have not done any math. I just seek to find a short way to calculate equivalent power
I had a problem with the implications of that equation. Power must be conserved so there are only certain values for p1, p2, p3 that can satisfy both that equation and the equation P = p1+p2+p3padmaraju bapala said:but how actually it be 1/P = 1/p1+1/p2+1/p3
I'd say the grad student who wrote that exercise is a nincompoop and so is the author of the book who didn't check up on him..sophiecentaur said:that equation is just a bit of Maths jiggery pokery and one shouldn't struggle too much with interpreting it in physical terms.
Shivansh Mathur said:I think i should put forward the question as under:
we have 2 identical resistors R(1) and R(2),
P(1) = Power of R(1)
P(2) = Power of R(2)
P(s) = Equivalent power of resistors R(1) and R(2) when connected in series.
P(s) = 1/P(1) + 1/P(2) HOW?
Ha Ha - you old dear!jim hardy said:Cyd Charisse
Let's say we have two bulbs of 100W-220V rating. We can get their resistance by R = V^2/P. We get 484 ohm.Shivansh Mathur said:I think i should put forward the question as under:
we have 2 identical resistors R(1) and R(2),
P(1) = Power of R(1)
P(2) = Power of R(2)
P(s) = Equivalent power of resistors R(1) and R(2) when connected in series.
P(s) = 1/P(1) + 1/P(2) HOW?
and if
P(p) = Equivalent power of resistors R(1) and R(2) when connected in parallel.
then,
P(p) = P(1) + P(2) HOW?
The following comment may have appeared higher up but here goes:ShoyebRC said:As, we get full power with parallel circuit and fraction of the full power with series circuit.
This keeps turning up in the thread. It can't be correct because it is dimensionally wrong.Shivansh Mathur said:P(s) = 1/P(1) + 1/P(2) HOW?
The equivalent power in series and parallel combination refers to the total power output of a circuit that is composed of multiple components connected in either series or parallel. It is used to simplify complex circuits and determine the overall power consumption or generation.
In series combination, the equivalent power is calculated by simply adding up the individual power values of each component in the circuit. This is because in a series circuit, the same current flows through each component, so the power is additive.
In parallel combination, the equivalent power is calculated by using the reciprocal of the individual power values and then adding them together. This is because in a parallel circuit, the voltage is the same across each component, so the power is inversely proportional to the resistance.
The main difference between equivalent power in series and parallel combination is the way it is calculated. In series, the power is additive, while in parallel, the power is inversely proportional to the resistance. Additionally, in series, the equivalent power is always less than the power of individual components, while in parallel, it can be greater.
Equivalent power is important in circuit analysis because it allows for the simplification of complex circuits and makes it easier to calculate the overall power consumption or generation. It also helps in understanding the behavior of different components in a circuit and how they affect the power output. Additionally, it is essential in designing and troubleshooting circuits to ensure they are functioning properly.